Solution

Stat 310B/Math 230B Theory of Probability
Homework 1 Solutions
Andrea Montanari
Due on 1/15/2014
Exercise [4.1.3]
We need to show that E(XIA ) = E(Y IA ), for all A ∈ G = σ(P). Let L = {A ∈ F : E(XIA ) = E(Y IA )}.
The assumption implies P ⊆ L. By Dynkin’s π − λ theorem, it suffices to show that L is λ-system, which
we proceed to check.
First, Ω ∈ L since Ω ∈ P ⊆ L. Second, if A ∈ L, B ∈ L and A ⊆ B, then taking the difference of the two
integral identities we see that B\A ∈ L. Finally, if Ai ∈ L, Ai ↑ A then applying dominated convergence we
conclude that A ∈ L.
Exercise [4.1.8]
1. It suffices to prove the claim for n = 2 as the general case then follows by n − 1 iterations. Fixing
hereafter n = 2, f (x1 , x2 ) = f1 (x1 )f2 (x2 ) is a finite valued element of mF+ (see Corollary 1.2.19),
hence νb = f µ is a measure on F (see Proposition 1.3.56), such that νb µ. Since νk = fk µk , if
A = A1 × A2 and Ak ∈ Fk , then by Fubini’s theorem and the definition of product measures,
ν(A) = ν1 (A1 )ν2 (A2 ) = µ1 (f1 IA1 )µ2 (f2 IA2 ) = µ(f IA ) = νb(A) .
In conclusion, the measures ν and νb coincide on the π-system R = {A1 × A2 : A1 ∈ F1 , A2 ∈ F2 }.
Further, since νk are σ-finite, there exist R` ∈ R such that ν(R` ) < ∞ and R` ↑ S. In view of the
remark following Proposition 1.1.39, ν = νb throughout F = σ(R), as claimed.
2. Consider the random variables Xk = fk (sk ) in probability space (S, F, µ). By definition σ(Xk ) consists
of all sets of the form Ak = {(s1 , . . . , sn ) : fk (sk ) ∈ Bk } for some Bk ∈ B. Therefore, ∩nk=1 Ak =
A1 × · · · × An and by the construction of product measure µ we have that
µ(
n
\
Ai ) = µ(A1 × · · · × An ) =
i=1
n
Y
µ(Ai )
i=1
(see Eq. (1.4.3)). This is precisely the definition of mutual independence of Xk and the same argument
applies in the probability space (S, F, ν).
Exercise [4.2.16]
1. Expanding both sides of the inequality, you see that it amounts to showing that
E[E(X|G1 )2 ] ≤ E[E(X|G2 )2 ]
Let Y = E(X|G1 ) and Z = E(X|G2 ). From Example 4.2.20 of the notes we know that both Y and Z
are square integrable. Further, by the tower property E(Z|G1 ) = Y so after also taking out what is
known
E(Y Z) = E(E(Y Z|G1 )) = E(Y E(Z|G1 )) = E(Y 2 ) .
Consequently,
0 ≤ E[(Z − Y )2 ] = E(Z 2 ) − 2E(ZY ) + E(Y 2 ) = E(Z 2 ) − E(Y 2 )
which is precisely what you are asked to prove. Note that this result is also a direct consequence of
Proposition 4.3.1.
1
2. Let Y = E(X|G). You get the stated identity by adding the equations
Var(Y ) = EY 2 − (EY )2 = EY 2 − (EX)2
and
E(Var(X|G)) = E(X − Y )2 = EX 2 − 2EXY + EY 2 = EX 2 − EY 2
Exercise [4.2.22]
1. Fix p > 0 and A ∈ G. Then, setting Y = |X|IA and Ux = P(|X| > x|G) we have from part (a) of
Lemma 1.4.31 and the definition of Ux that
Z ∞
E[|X|p IA ] = EY p =
py p−1 P(Y > y)dy
0
Z ∞
Z ∞
p−1
px E[I{|X|>x} IA ]dx =
pxp−1 E[Ux IA ]dx .
=
0
0
Recall that Ux IA ∈ mG for each x ≥ 0. Without loss of generality we further assume that the nonnegative function h(x, ω) = pxp−1 Ux IA is measurable on the product space B × G (the easiest way
to see this is by taking the version of Ux given by the measure of the open interval (x, ∞) under the
R.C.P.D. of |X| given G, which exists by Proposition 4.4.3).
R∞
Thus, by Fubini’s theorem Z = 0 pxp−1 Ux dx is measurable on G and
E[|X|p IA ] =
Z
∞
Z
E[h(x, ω)]dx = E[
0
∞
h(x, ω)dx] = E[ZIA ] .
0
In particular, EZ = E|X|p is finite and as the preceding applies for all A ∈ G, it follows by definition
of conditional expectation that Z = E[|X|p |G].
2. Fixing a > 0 let Va = P(|X| ≥ a|G). By the monotonicity of the conditional expectation Ux ≥ 0 for
any x ≥ 0 and further Ux ≥ Va whenever x ∈ [0, a). Hence, for any a > 0 and A ∈ G we have from our
proof of part (a) that
Z ∞
Z a
p
p−1
E[|X| IA ] =
px E[Ux IA ]dx ≥
pxp−1 dxE[Va IA ] = ap E[Va IA ] .
0
0
To conclude that almost surely Va ≤ a−p E[|X|p |G] consider the above inequality for An = {ω : ap Va ≥
n−1 + E[|X|p |G] }, then take n → ∞.
Exercise [4.2.21]
1. Taking out what is known,
E(XZ) = E(E(XZ|G)) = E(ZE(X|G)) = EZ 2 .
Therefore,
E(X − Z)2 = EX 2 − 2EXZ + EZ 2 = EX 2 − EZ 2 = 0 ,
from which we deduce that Z = X a.s.
2. We know from (cJENSEN) that almost surely |Z| = |E(X|G)| ≤ E(|X||G). Hence, if P(|Z| <
E(|X||G)) > 0 then E(|E(X|G)|) < E(E(|X||G)) = E(|X|), in contradiction with our hypothesis
that |Z| = |E(X|G)| has the same law as |X| (hence the same expected value). We thus conclude that
2
|Z| = E(|X||G) almost surely. Note that A = {Z ≥ 0} is by definition of Z in G and further by the
preceding,
E[XIA ] = E[ZIA ] = E[|Z|IA ] = E[E(|X| |G)IA ] = E[|X|IA ] .
That is, E[(|X| − X)IA ] = 0, namely, X ≥ 0 for almost every ω ∈ A. Our hypothesis that E[X|G]
has the same law as X implies that the same hypothesis holds for Y = X − c and any non-random
constant c. Therefore, by the preceding we get that
P({X < c ≤ E(X|G)}) = P({Y < 0 ≤ E(Y |G)}) = 0.
S
Since {X < E(X|G)} = c∈Q {X < c ≤ E(X|G)}, it follows that X ≥ E(X|G) a.s. To complete the
proof re-run the above argument for −X instead of X.
Exercise [4.2.23]
For ε ≥ 0, let Uε = E(|X|p |G) + ε]1/p in Lp (Ω, G, P) and Vε = E(|Y |q |G) + ε]1/q , in Lq (Ω, G, P), that per
ε > 0 are both uniformly bounded below away from zero. Recall that
yq
xp
+
− xy ≥ 0,
p
q
for all x, y ≥ 0
(which you verify by considering the first two derivatives in x
each ω and ε > 0,
X(ω)Y (ω) 1 X(ω) p
≤ +
Uε (ω)Vε (ω)
p Uε (ω)
of the function on the left side). Hence, for
q
1 Y (ω) .
q Vε (ω)
With 1/Uε and 1/Vε uniformly bounded, the expectation of both sides conditional upon G is well defined,
and it follows from monotonicity of the C.E. (i.e. Corollary 4.2.6), upon taking out what is known that for
any ε > 0 and a.e. ω,
1 U0p
1 V0q
1 1
E(|XY | |G)
≤
+ = 1.
p +
q ≤
Uε Vε
p Uε
q Vε
p q
Multiplying both sides by Uε Vε and considering εk ↓ 0 yields the stated claim that E(|XY | |G) ≤ U0 V0 .
Exercise [4.2.27]
(a) implies (b): (b) holds for indicator functions h1 and h2 by (a). By linearity of E(·|G) and (cDOM), upon
using the standard machine (see Definition 1.3.6), we see that (b) holds for all bounded Borel functions h1
and h2 .
(b) implies (c): We know that G ⊆ H = σ(G, σ(X2 )) and have to show that for any A ∈ H,
E(E(h1 (X1 )|G)IA ) = E(h1 (X1 )IA ).
We use Dynkin’s π-λ theorem for the generating class {A = B ∩ C : B ∈ G, C ∈ σ(X2 )} of H, which is
closed under finite intersection. Then, by (b),
E[h1 (X1 )IB IC ] = E[E(h1 (X1 )IC |G)IB ] = E[E(h1 (X1 )|G)E(IC |G)IB ] .
Thus, using the tower property and taking out the G-measurable IB E(h1 (X1 )|G),
E{E(h1 (X1 )|G)IA } = E{E[E(h1 (X1 )|G)IB IC |G]} = E(h1 (X1 )IA ).
(c) implies (a): By the tower property,
P(X1 ∈ B1 , X2 ∈ B2 |G) = E[E(IB1 (X1 )IB2 (X2 )|H) | G].
3
As IB2 (X2 ) is H-measurable, using (c) for h(x) = IB1 (x) and taking out the G-measurable E(IB1 (X1 )|G) we
also have that
E[E(IB1 (X1 )IB2 (X2 )|H) | G]
= E[IB2 (X2 )E(IB1 (X1 )|H) | G]
= E[IB2 (X2 )E(IB1 (X1 )|G) | G] =
Y
i=1,2
4
P(Xi ∈ Bi |G).

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